All categories

3 years ago

. Calculate the kinetic energy of a 100.0-kg meteor approaching the Earth at a speed of 10.0 km/s. Remember that 1 km = 1000 m.

Physics

2 answers:

anastassius [24]3 years ago

7 0

Ke = (1/2)mv²

m = 100kg, v = 10 km/s = 10*1000 = 10000m/s

Ke = (1/2)*100*10000

Ke = 500000 Joules

Elan Coil [88]3 years ago

4 0

Kinetic energy is solved by the formula,
KE = 0.5 x m x v²
First, convert the given speed in km/s to that with unit of m/s.
(10 km/s) x (1000 m/ 1 km) = 10000 m/s
Use the formula to obtain the KE
KE = 0.5 x (100 kg) x (10000 m/s )²
Thus, the value of its kinetic energy is 5x10^9 J.

You might be interested in

1) If you release a rubber band that had 10 units of elastic energy, 12 units of movement energy cannot be produced. Why not?

Mila [183]

Answer:

press dat crown for meh

Explanation:

4 0

2 years ago

Read 2 more answers

Question 6

Margaret [11]

Limestone and dolomite are the rocks present in the locations which leads to the formation of caves.

<h2>Formation of caves</h2>

The type of rocks that once existed in these locations are limestone and dolomite whereas the pH of the nearby groundwater is slightly acidic which is responsible for the formation of caves. Caves are formed by the dissolution of limestone due to acid rain.

Rainwater reacts with carbon dioxide from the air and percolates through the soil, which turns into a weak acid. This slowly dissolves out the limestone which become turn to form caves so we can conclude that Limestone and dolomite are the rocks present in the locations which leads to the formation of caves.

Learn more about caves here: brainly.com/question/7965722

Learn more: brainly.com/question/26111031

4 0

2 years ago

The work that a force does by acting on an object is equal to what?

maw [93]

2nd and only 2nd option is right

3 0

3 years ago

How does the sun’s gravitational attraction impact Earth’s motion?

Tasya [4]

The Sun's gravitational pull keeps our planet orbiting the Sun <span>in a nice nearly-circular orbit.</span>

7 0

3 years ago

A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.

kap26 [50]

Answer:

The frequencies are $(f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)$

Explanation:

From the question we are told that

The length of the ear canal is $l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m$

The speed of sound is assumed to be $v_s = 344 \ m/s$

Now taking look at a typical ear canal we see that we assume it is a closed pipe

Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

$f = \frac{v_s}{4 * l }$

substituting values

$f = \frac{344}{4 * 0.013 }$

$f = 6615.4 \ Hz$

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

$f_1 = \frac{3v_s}{4 * l}$

substituting values

$f_1 = \frac{3 * 344}{4 * 0.013}$

$f_1 = 19846.2 \ Hz$

Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies

6 0

3 years ago